Suppose $k$ is a ring and $R$ is the ring of $2\times 2$ upper triangular matrices with entries in $k$. Via left matrix multiplication, $k^2$ viewed as column vectors becomes a left $R$-module. Now it turns out that $k^2$ is projective as a left $R$-module. Today we'll see one methods to show this. Before we do this, let us briefly outline another solution to this problem that will make the reader appreciate the second method. The most straightforward approach to this problem is perhaps using the lifting criterion, or at least the following variation of it: define a surjection $R\to k^2$ by $1\mapsto [0,1]^t$. Now, one just has find a splitting of this surjection. Exercise: find a map that's a splitting and prove carefully that it is actually an $R$-module homomorphism.

Now let's see another way to do this. Instead of constructing a splitting, we could continue the surjection $R\to k^2$ defined by $1\mapsto [0,1]^t$ above into a projective resolution. Since maps out of free $R$-modules are easy to construct, if the resoluting projective resolution is small, then it will be easy to see by inspecting the resolution whether $k^2$ is projective (or so we hope).

To continue the projective resolution, we just find the kernel of $R\to k^2$ and map onto it with another free module. The reader may wish to write this process down before reading on.
Doing this, we get a free resolution as follows. If we denote by $\alpha:R\to R$ the map $1\mapsto\left(\begin{smallmatrix}1 & 0\\ 0 & 0\end{smallmatrix}\right)$ and by $\beta:R\to R$ the map $1\mapsto\left(\begin{smallmatrix}0 & 0\\ 0 & 1\end{smallmatrix}\right)$, our resolution is
$$\cdots R\xrightarrow{\beta}R\xrightarrow{\alpha}R\xrightarrow{\beta}R\xrightarrow{\alpha}R\to k^2\to 0$$
Can you see why $k$ has to be projective? If $M$ is any $R$-module, then applying $\mathrm{Hom}_R(-,M)$ to the resolution part gives a complex of abelian groups $0\to M\xrightarrow{\alpha^*} M\xrightarrow{\beta^*} M\xrightarrow{\alpha^*}\cdots$ where $\alpha^*$ is multiplication by $\left(\begin{smallmatrix}1 & 0\\ 0 & 0\end{smallmatrix}\right)$ and $\beta^*$ is multiplication by $\left(\begin{smallmatrix}0 & 0\\ 0 & 1\end{smallmatrix}\right)$. Since these two matrices sum to the identity, this complex is still exact! Hence, $k^2$ is projective.